In case you didn’t encounter it on Facebook, here is an excellent logic puzzle from Singapore:
Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates.
May 15, May 16, May 19
June 17, June 18
July 14, July 16
August 14, August 15, August 17
Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively, so Albert knows the month while Bernard knows the day.
Albert: I don’t know when Cheryl’s birthday is, but I know that Bernard does not know too.
Bernard: At first I didn’t know when Cheryl’s birthday is, but I know now.
Albert: Then I also know when Cheryl’s birthday is.
1) Albert says “I don’t know when Cheryl’s birthday is, but I know that Bernard does not know too.” If the birthday had been May 19 or June 18, then Bernard would have known it immediately after being told the day, because those days only appear once. In order for Albert to know this about Bernard’s knowledge, he must know that the month is not May or June. (Becuase if the month was either may or June, Bernard might know the birthday right away). Therefore the birthday is not in May or June.
2) Bernard says: “At first I didn’t know when Cheryl’s birthday is, but I know now.” After Albert gives Bernard the information that it is either July or August, Bernard learns the exact date. This indicates that the day is not the 14th, because the 14th is present in both July and August. In order for Bernard’s information to not remain ambiguous, it must be one of July 16, August 15, or August 17, because those three have a unique day value out of the remaining possibilities.
3) Albert says: “Then I also know when Cheryl’s birthday is.” This indicates that for our three remaining possibilities, the month is not ambiguous, and therefore it must be July. (Because August had two possibilities left, while July had one). Therefore the birthday is on July 16.
I found it fairly easy, but I am experienced with logic puzzles, including those where characters in the puzzle use their knowledge about the knowledge of others in the puzzle in order to solve it. If someone wasn’t experienced with these kinds of logic puzzles it would probably seem very hard. (The first time I encountered one of these I think I agonized over it for many hours).
characters in the puzzle use their knowledge about the knowledge of others in the puzzle in order to solve it
This is known as a “common knowledge” puzzle. For an example of a much more difficult (IMO) puzzle involving common knowledge, take a look at this one. (Although my admission here that the puzzle involves common knowledge might actually make it significantly easier.)
Can it be solved using probabilities? I mean, if, for example, the Guru says ‘I see nobody on this island who has blue eyes’, then the whole 100 brown-eyed people can pack and leave at once. If the Guru says ‘I see at least one blue-eyed person’ AND thereare two BE present, neither of them leaves that night and so next morning they both know they are BE and can leave the following night, and so on?
I mean, the ‘perfect logicians’ part put me into thinking like ‘...and if there are 3 BlE and 100 BrE, and the Guru says ‘I see at least 1 BlE’, then at that moment each one of the three BlE thinks there’s 2⁄3 chances she means either of the other two, so next time just before noon, when they converge again, one of the three BlE finds the other two and goes away without saying anything. Then if next morning the two others are found to have left, he knows he’s chance of being the only BlE left has gone up and presents himself for inspection. When he is confirmed as BlE, next time it automatically releases all BrE. Now let’s consider the case of four BlE present...′ etc. I doubt it can be done easily for large groups of people, though, unless they cooperate. The easiest way to do it is to appear before the Guru in pairs:)
A census taker approaches a woman leaning on her gate and asks about her children. She says, “I have three children and the product of their ages is seventy–two. The sum of their ages is the number on this gate.” The census taker does some calculation and claims not to have enough information. The woman enters her house, but before slamming the door tells the census taker, “I have to see to my eldest child who is in bed with measles.” The census taker departs, satisfied.
What are the ages of the children?
I actually remember finding blue eyes easier than three children, but I don’t remember my relative ages when I first heard them, and I discussed both with my family at the time. But blue eyes had an “obvious” approach which worked after routine application, and three children/Cheryl require you to extract a different insight from every step of the puzzle.
I found it easy enough to solve (the same way you did), and I’m not very experienced with logic puzzles. Solving it instead of giving up and looking up the answer made me feel good!
For Bernard to be unable to determine Cheryl’s birthday upon being told the day, the day must be insufficient to specify the month. In other words, the day has to be one of the numbers that appears more than once in the list. This immediately rules out 18 and 19, which both only appear once. Moreover, for Albert to know that Bernard doesn’t know Cheryl’s birthday, the month he was given must not contain either 18 or 19 as a possible day; otherwise, it would have been possible for Bernard to figure out the month from the date, and Albert could not know that Bernard did not know Cheryl’s birthday. This rules out May (which contains 19) and June (which contains 18).
Upon hearing that Albert knew he did not know Cheryl’s birthday, Bernard would gain the above information, and know that Cheryl’s birthday falls in either July or August. This means that the information he was given must be sufficient to discriminate between these two months, i.e. whatever the day Cheryl gave him, it cannot appear in both months. This rules out 14. The remaining possibilities are July 16, August 15, and August 17.
This is where I got stuck. There doesn’t seem to be any more information in the problem that would allow further discrimination between these three possibilities. Moreover, this makes Albert’s assertion that he now knows Cheryl’s birthday after hearing Bernard absurd; how could he possibly know which month it is?
I’m still unsure how to proceed right now. I’ll give it ten or so more minutes of thought, and if I fail to come up with anything after that, I’ll look at the answer.
EDIT: Man, I feel stupid. The answer came to me right after I commented, and it turns that my mistake was that I had unconsciously conflated Albert with the reader. The reader doesn’t know the month, and therefore without further information, it’s impossible to determine which of the three possibilities Cheryl’s birthday actually falls on. However, Albert does know the month, and whichever one it is, it cannot contain more than a single possible day for Cheryl’s birthday (because if it did, Albert wouldn’t be able to tell which day it was). Of the set of possibilities I had originally {July 16, August 15, August 17}, the month of August contains two possible days—so if Cheryl’s birthday were in August, Albert would not know whether it was August 15 or August 17. Therefore, Cheryl’s birthday must fall in July, and so the answer is:
July 16
(META: That was pretty fun! We should do this more often.)
Raymond Smullyan calls these sorts of puzzles (where characters’ ability to solve the puzzle is used by the reader to solve the puzzle) “metapuzzles”. There are some more examples in his books.
In case you didn’t encounter it on Facebook, here is an excellent logic puzzle from Singapore:
Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates.
May 15, May 16, May 19
June 17, June 18
July 14, July 16
August 14, August 15, August 17
Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively, so Albert knows the month while Bernard knows the day.
Albert: I don’t know when Cheryl’s birthday is, but I know that Bernard does not know too.
Bernard: At first I didn’t know when Cheryl’s birthday is, but I know now.
Albert: Then I also know when Cheryl’s birthday is.
When is Cheryl’s birthday?
Poll:
How long did it take you to solve this problem?
[pollid:853]
SPOILER ALERT:
Here is the solution I just came up with.
1) Albert says “I don’t know when Cheryl’s birthday is, but I know that Bernard does not know too.”
If the birthday had been May 19 or June 18, then Bernard would have known it immediately after being told the day, because those days only appear once. In order for Albert to know this about Bernard’s knowledge, he must know that the month is not May or June. (Becuase if the month was either may or June, Bernard might know the birthday right away). Therefore the birthday is not in May or June.
2) Bernard says: “At first I didn’t know when Cheryl’s birthday is, but I know now.”
After Albert gives Bernard the information that it is either July or August, Bernard learns the exact date. This indicates that the day is not the 14th, because the 14th is present in both July and August. In order for Bernard’s information to not remain ambiguous, it must be one of July 16, August 15, or August 17, because those three have a unique day value out of the remaining possibilities.
3) Albert says: “Then I also know when Cheryl’s birthday is.” This indicates that for our three remaining possibilities, the month is not ambiguous, and therefore it must be July. (Because August had two possibilities left, while July had one). Therefore the birthday is on July 16.
I found it fairly easy, but I am experienced with logic puzzles, including those where characters in the puzzle use their knowledge about the knowledge of others in the puzzle in order to solve it. If someone wasn’t experienced with these kinds of logic puzzles it would probably seem very hard. (The first time I encountered one of these I think I agonized over it for many hours).
Instead of writing spoiler alert using rot13 is better.
This is known as a “common knowledge” puzzle. For an example of a much more difficult (IMO) puzzle involving common knowledge, take a look at this one. (Although my admission here that the puzzle involves common knowledge might actually make it significantly easier.)
Can it be solved using probabilities? I mean, if, for example, the Guru says ‘I see nobody on this island who has blue eyes’, then the whole 100 brown-eyed people can pack and leave at once. If the Guru says ‘I see at least one blue-eyed person’ AND thereare two BE present, neither of them leaves that night and so next morning they both know they are BE and can leave the following night, and so on?
The approach you describe is sensible, but I don’t see what it has to do with probabilities; all the probabilities involved are either 0 or 1.
I mean, the ‘perfect logicians’ part put me into thinking like ‘...and if there are 3 BlE and 100 BrE, and the Guru says ‘I see at least 1 BlE’, then at that moment each one of the three BlE thinks there’s 2⁄3 chances she means either of the other two, so next time just before noon, when they converge again, one of the three BlE finds the other two and goes away without saying anything. Then if next morning the two others are found to have left, he knows he’s chance of being the only BlE left has gone up and presents himself for inspection. When he is confirmed as BlE, next time it automatically releases all BrE. Now let’s consider the case of four BlE present...′ etc. I doubt it can be done easily for large groups of people, though, unless they cooperate. The easiest way to do it is to appear before the Guru in pairs:)
Another one that seems of similar difficulty to Cheryl’s birthday is http://en.wikipedia.org/wiki/Ages_of_Three_Children_puzzle (the article contains spoilers):
What are the ages of the children?
I actually remember finding blue eyes easier than three children, but I don’t remember my relative ages when I first heard them, and I discussed both with my family at the time. But blue eyes had an “obvious” approach which worked after routine application, and three children/Cheryl require you to extract a different insight from every step of the puzzle.
I found it easy enough to solve (the same way you did), and I’m not very experienced with logic puzzles. Solving it instead of giving up and looking up the answer made me feel good!
(Posted without looking at the replies.)
For Bernard to be unable to determine Cheryl’s birthday upon being told the day, the day must be insufficient to specify the month. In other words, the day has to be one of the numbers that appears more than once in the list. This immediately rules out 18 and 19, which both only appear once. Moreover, for Albert to know that Bernard doesn’t know Cheryl’s birthday, the month he was given must not contain either 18 or 19 as a possible day; otherwise, it would have been possible for Bernard to figure out the month from the date, and Albert could not know that Bernard did not know Cheryl’s birthday. This rules out May (which contains 19) and June (which contains 18).
Upon hearing that Albert knew he did not know Cheryl’s birthday, Bernard would gain the above information, and know that Cheryl’s birthday falls in either July or August. This means that the information he was given must be sufficient to discriminate between these two months, i.e. whatever the day Cheryl gave him, it cannot appear in both months. This rules out 14. The remaining possibilities are July 16, August 15, and August 17.
This is where I got stuck. There doesn’t seem to be any more information in the problem that would allow further discrimination between these three possibilities. Moreover, this makes Albert’s assertion that he now knows Cheryl’s birthday after hearing Bernard absurd; how could he possibly know which month it is?
I’m still unsure how to proceed right now. I’ll give it ten or so more minutes of thought, and if I fail to come up with anything after that, I’ll look at the answer.
EDIT: Man, I feel stupid. The answer came to me right after I commented, and it turns that my mistake was that I had unconsciously conflated Albert with the reader. The reader doesn’t know the month, and therefore without further information, it’s impossible to determine which of the three possibilities Cheryl’s birthday actually falls on. However, Albert does know the month, and whichever one it is, it cannot contain more than a single possible day for Cheryl’s birthday (because if it did, Albert wouldn’t be able to tell which day it was). Of the set of possibilities I had originally {July 16, August 15, August 17}, the month of August contains two possible days—so if Cheryl’s birthday were in August, Albert would not know whether it was August 15 or August 17. Therefore, Cheryl’s birthday must fall in July, and so the answer is:
July 16
(META: That was pretty fun! We should do this more often.)
Perhaps a monthly puzzle thread.
Raymond Smullyan calls these sorts of puzzles (where characters’ ability to solve the puzzle is used by the reader to solve the puzzle) “metapuzzles”. There are some more examples in his books.
August 17?
Edit: I made the mistake of discarding June but not May.
SPOILER: here is why it’s not Aug 17.